There are square roots, so we need the expressions inside the roots to be non-negative:

x\leq 0\text{ and }\sin \sqrt{x}\leq 0

Let’s look at the second inequality. The sin function is positive in the first half of its cycle, meaning when the following holds:

0\leq \sqrt{x}\leq \pi

But sin function has endless cycles. Let’s express this with parameter k (integer). Adding the cycle in both sides will result in

2\pi k \leq \sqrt{x}\leq \pi+2\pi k

2\pi k \leq \sqrt{x}\leq \pi (1+2k)

4\pi^2 k^2 \leq x\leq \pi^2(1+2k)^2

Note that since x is between two quadratic expressions, x is necessarily positive or zero. That is, the first inequality also exists (had it not happened, we would have to intersect them). Therefore, that’s the final answer.

Have a question? Found a mistake? – Write a comment below! Was it helpful? You can buy me a cup of coffee here, which will make me very happy and will help me upload more solutions!